\(x^2+2xy+6x+6y+2y^2+8=0\\ \Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+y^2=-8\)

Ta có \(y^2\ge0\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)\le-8\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9\le1\\ \Leftrightarrow\left(x+y+3\right)^2\le1\\ \Leftrightarrow\left|x+y+3\right|\le1\\ \Leftrightarrow-1\le x+y+3\le1\\ \Leftrightarrow2012\le B\le2014\)

\(B_{min}=2012\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2012\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

\(B_{max}=2014\Leftrightarrow\left\{{}\begin{matrix}x+y+2016=2014\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

đề bài sai r bn ơi phải là +10 chứ ko phải +8 đâu nhá

Mình biết hơi muộn

\(A=x^2+2xy+6x+6y+2y^2+8\Leftrightarrow x^2+2xy+6x+6y+y^2+9-1\)

\(A=0\Rightarrow\left(x+y+3\right)^2+y^2-1=0\)

\(\Rightarrow-1\le x+y+3\le1\) .

\(\Rightarrow2012\le x+y+3+2013\le2014\)

\(\Rightarrow2012\le B\le2014\)

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)

\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)

Ta thấy : \(1-y^2\le1\forall y\) \(\Rightarrow\left(x+y+3\right)^2\le1\)

\(\Rightarrow-1\le x+y+3\le1\)

\(\Rightarrow-1+2013\le x+y+3+2013\le1+2013\)

\(\Rightarrow2012\le x+y+2016\le2014\)

Vậy ta có : 

+) Min \(B=2012\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-4\end{cases}}\)

+) Max \(M=2014\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

We have:

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow\left(x+y+3\right)^2=-y^2+1\)

\(\Rightarrow\left(x+y+3\right)^2\le1\)

\(\Rightarrow-1\le x+y+3\le1\)

\(\Rightarrow2015\le x+y+2019\le2017\)

Sign '=' happen when \(x=-4;x=-2;y=0\)